For 2≤r≤n, Crn+2Cr+1n+Cr+2n=
Cr-1n+1
2Cr+1n+1
2Crn+2
Cr+2n+2
Finding the value:
Given, r&n are positive integer, 2≤r≤n
We know that,
Crn+Cr+1n=Cr+1n+1
Using the same for given equation,
Crn+2Cr+1n+Cr+2n=Crn+Cr+1n+Cr+1n+Cr+2n=Cr+1n+1+Cr+2n+1=Cr+2n+2
Hence the correct option is (D)
Let r and n be positive integers such that 1≤r≤n. Then prove the following :
(i) nCrnCr−1=n−r+1r (ii) nn−1Cr−1=(n−r+1)nCr−1 (iii) nCrn−1Cr−1=nr (iv) nCr+2nCr−1+nCr−2=n+2Cr
nCr+2nCr−1+nCr−2 =